Declaring a long long array with unsigned int size

Question

I was doing a problem when I encountered a segmentation fault while declaring an array as:

long long ways[max+1];

where,

unsigned int max = findMax(l,T); // l is an unsigned int array and T is an int.

and findMax is a function of the type:

unsigned int findMax(unsigned int arr[],int size)

How can I resolve this?

How big is `max`? Probably stack overflow. – Yu Hao Dec 19 '14 at 12:18 — Yu Hao, Dec 19 '14 at 12:18

Klas Lindbäck · Accepted Answer · 2014-12-19T13:27:51.640

3

A probable reason is that the array is too big for the stack.

Typical stack sizes as of now is 1-16 Mb (could be considerably less in an embedded system). If long long is 8 bytes it would mean that allocating one array of more than 125000 elements could be problematic. And you also want to leave room for other auto variables.

Really big arrays should not be allocated on the stack.

You could try allocating it in the heap instead:

long long *ways = calloc(max+1, sizeof *ways);
if (ways == NULL) {
    // allocation failed!!
}
// Do stuff.
free(ways);

edited Dec 19 '14 at 13:27

answered Dec 19 '14 at 12:30

Klas Lindbäck

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Minor idea: `long long *ways = calloc(max+1, sizeof *ways);` style instead. IMO, easier to code and maintain. – chux - Reinstate Monica Dec 19 '14 at 13:13
1

@chux Agreed! That way there is one less place to change if you want to change the type the array. – Klas Lindbäck Dec 19 '14 at 13:29
I'm sorry, I cannot spot the difference between these two. – Chandan Dec 19 '14 at 13:58
@Chandan I have updated my answer. `sizeof *ways` is better than `sizeof (long long)`. – Klas Lindbäck Dec 19 '14 at 14:51
It will use the same amount of memory, but from the heap instead of from the stack. – Klas Lindbäck Dec 20 '14 at 21:31

Declaring a long long array with unsigned int size

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